Conditional probabilities: Finding the p.m.f. of $\mathbb{E}[X\mid Y]$

Question

Let $X, Y$ be two r.v. with associated joint p.m.f.

\begin{array}{|c|c|c|c|} \hline & & Y & \\ \hline & & 0 & 1 \\ \hline X & 0 & 3/10 & 2/10 \\ \hline & 1 & 1/10 & 4/10 \\ \hline \end{array}

1. What values can $\mathbb{E}[X\mid Y]$ take?

2. Compute the p.m.f. of $\mathbb{E}[X\mid Y]$

My Attempt

1. The formula for the expectation of conditional probabilities is $$\mathbb{E}[X\mid Y=y]=\sum_{x\in X(\Omega)}xp_{X\mid Y}(x\mid y)=\sum_{x\in X(\Omega)}x\frac{p_{X,Y}(x,y)}{p_Y(y)}$$ so the possible values are $\mathbb{E}[X\mid Y=0]=1(\frac{1/10}{4/10})=1/4$ and $\mathbb{E}[X\mid Y=1]=1\frac{4/10}{6/10}=2/3$

2. I'm confused on this part because I have never tried finding the p.m.f. for an expectation. How should I go about this problem?

Answer 1

Notice that the conditional expectation is in fact a random variable. And for your example you just give Y a value to get a certain conditional expectation, so instead of an exact number you can just using Y to get the Pmf.

Answer 2

Combining the facts that $\mathbb E\left[X\mid Y=0\right]=\frac14$ and $\mathbb E\left[X\mid Y=1\right]=\frac23$, you can say that

$$\mathbb E\left[X\mid Y=y\right]=\left(\frac23-\frac14\right)y+\frac14=\frac{5y}{12}+\frac14 \quad,\,\,y\in \{0,1\}$$

This means $\mathbb E\left[X\mid Y\right]$ is a random variable given by

$$\mathbb E\left[X\mid Y\right]=\frac{5Y}{12}+\frac14 \quad,\,\text{with probability }1$$

So $\mathbb E\left[X\mid Y\right]$ takes the value $\frac14$ with probability $\mathbb P(Y=0)$ and the value $\frac23$ with probability $\mathbb P(Y=1)$.